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Question

Solve: $$\displaystyle { \tan }^{ -1 }\frac { 1-x }{ 1+x } =\frac { 1 }{ 2 } { \tan }^{ -1 }x,\left( x>0 \right) $$


Solution

$$\displaystyle  { \tan }^{ -1 }\frac { 1-x }{ 1+x } =\frac { 1 }{ 2 } { \tan }^{ -1 }x$$

$$\displaystyle\Rightarrow { \tan }^{ -1 }1-{ \tan }^{ -1 }x=\frac { 1 }{ 2 } { \tan }^{ -1 }x$$

$$\displaystyle \Rightarrow \frac { \pi  }{ 4 } =\frac { 3 }{ 2 } { \tan }^{ -1 }x$$

$$\displaystyle \Rightarrow { \tan }^{ -1 }x=\frac { \pi  }{ 6 } $$

$$\displaystyle \Rightarrow x=\tan { \frac { \pi  }{ 6 }  } $$

$$\displaystyle \therefore x=\frac { 1 }{ \sqrt { 3 }  } $$

Mathematics
RS Agarwal
Standard XII

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